An aqueous solution containing `Na^(+),Sn^(2+),Cl^(-)` & `SO_(4)^(2-)` ions, all at unit concentration is electrolysed between a silver anode and a platium cathode. What changes occur at the electrodes when curret is passed through the cell? Given `E_(Ag^(+)|Ag)^(0)=0.799V`,
`E_(Sn^(2+)|Cn)^(0)=-0.14V,E_(Cl_(2)|Cr^(-))^(0)=1.36V,E_(S_(2)O_(8)^(2-)|SO_(4)^(2-))=2V,E_(Sn^(4+)|Sn^(2+))^(0)=0.13V`
(A). `Sn^(2+)` is reduced and `Cl^(-)` is oxidized
(B). Ag is oxidized and `Sn^(2+)` is reduced
(C). `Sn^(2+)` is reduced and `Sn^(2+)` is oxidized
(D). `H^(+)` is reduced and `Sn^(2+)` is oxidised

Explanation

At anode either Ag can oxidised to `Ag^(+)` or `Sn^(2+)` to `Sn^(4+)` or `Cl^(-)` to `Cl_(2)` or `SO_(4)^(2-)` to `S_(2)O_(8)^(2-)` Their respective oxidation potential values are `-0.799V,0.13V,-1.36V` and `-2V`. From these values it is evident that `Sn^(2+)` would be oxidised first, followed by Ag at anode. At cathode, either `Na^(+)` can get reduced to `Na` or `Sn^(2+)` or `Sn^(2+)` to `H^(+)` or `H_(2)`. The reduction potential value for `Na^(+)` is highly negative while for `Sn^(2+)|Sn` is `-0.14V` and for `H^(+)+e^(-)to1//2H_(2)(E_(H^(+)|H_(2))=0.059log(1)/(10^(-7)))` is `-0.413V`. Thus `Sn^(2+)` will get reduced at cathode followed by `H^(+)therefore(C)`